Mathematics - Vector Question with Solution | TestHub

Let the plane be$\mathrm{a}(\mathrm{x}-1)+\mathrm{b}(\mathrm{y}-0)+\mathrm{c}(\mathrm{z}-0)=0$

$$\Rightarrow \mathrm{ax}+\mathrm{by}+\mathrm{cz}-\mathrm{a}=0$$

Pass through$(0,1,0)\Rightarrow \mathrm{a}.0+\mathrm{b}.1+\mathrm{c}.0-\mathrm{a}=0$

$$\Rightarrow \mathrm{a}=\mathrm{b}$$

Normal to plane :$\mathrm{a}\hat{\mathrm{i}}+\mathrm{b}\hat{\mathrm{j}}+\mathrm{c}\hat{\mathrm{k}}$ Makes angle$\frac{\pi }{4}$ with$\hat{\mathrm{i}}+\hat{\mathrm{j}}$

$$\begin{array}{rlr}& \therefore \frac{a+b}{\sqrt{a^2+b^2+c^2}\cdot \sqrt{2}}=\frac{1}{\sqrt{2}}& \Rightarrow (a+b{)}^2=a^2+b^2+c^2\end{array}$$

By (i) and (ii) :$4{\mathrm{a}}^2=2{\mathrm{a}}^2+{\mathrm{c}}^2\Rightarrow 2{\mathrm{a}}^2={\mathrm{c}}^2$

$$\begin{array}{rlr}& \Rightarrow \sqrt{2}a=c;-\sqrt{2}a=c& \therefore a:b:c=1:1:\sqrt{2}\end{array}$$